Problem 36 - Contemporary Precalculus

Question

Find the exact functional value without using a calculator. $$\tan ^{-1}[\tan (-4 \pi / 3)]$$

Step-by-Step Solution

Answer

Answer: The exact functional value is \(\frac{\pi}{6}\).

Step 1: Evaluate the inner tangent function

First, we need to evaluate \(\tan(-4 \pi / 3)\). We know that because tangent is a periodic function with a period of \(\pi\), the tangent of an angle remains the same when you add or subtract integer multiples of \(\pi\). So, $$\tan(-4 \pi / 3) = \tan(-4 \pi / 3 + 2\pi)$$

Step 2: Add 2π to angle

Now, add \(2\pi\) to the angle: $$\tan(-4 \pi / 3 + 2\pi) = \tan( 2 \pi / 3)$$

Step 3: Evaluate the new tangent function

Now, we evaluate \(\tan(2\pi/3)\). Notice that this is in the second quadrant, where tangent is positive. We can use the co-function identity: $$\tan(\pi/2 - \theta) = \cot(\theta)$$ So, $$\tan(2\pi/3) = \tan(\pi/2 - \pi/3) = \cot(\pi/3)$$

Step 4: Evaluate the cotangent function

Next, let's evaluate \(\cot(\pi/3)\) and recall that \(\cot(\theta) = 1/\tan(\theta)\). We know that \(\tan(\pi/3) = \sqrt{3}\), so $$\cot(\pi/3) = \frac{1}{\sqrt{3}}$$

Step 5: Find the angle for inverse tangent

Now that we have evaluated the inner tangent function, the original expression becomes: $$\tan^{-1}[\tan(2\pi/3)] = \tan^{-1}\left(\frac{1}{\sqrt{3}}\right)$$ Since the tangent function is odd, we know that \(\tan^{-1}(\theta) = -\tan^{-1}(-\theta)\). In this case, with \(\theta = \pi/6\), we have $$\tan^{-1}(\frac{1}{\sqrt{3}}) = \pi/6$$

Step 6: Final functional value

Finally, we can write the exact functional value without using a calculator: $$\tan ^{-1}[\tan (-4 \pi / 3)] = \frac{\pi}{6}$$

Key Concepts

Periodic Functions

Periodic functions repeat their values at regular intervals. For the tangent function, this interval, or period, is \( \pi \). This means that for any angle \( \theta \), \( \tan(\theta) = \tan(\theta + n\pi) \) for any integer \( n \). This property is useful for simplifying expressions by reducing large or negative angles to equivalent angles within a more manageable range.

In our problem, \( \tan(-4\pi/3) \) can be simplified using this periodicity by adding \( 2\pi \) (which is equivalent to adding \( \pi \) twice), resulting in \( \tan(2\pi/3) \). This simplification helps us evaluate the function more easily.

Tangent Function

The tangent function, \( \tan(\theta) \), is the ratio of the sine and cosine functions, defined as \( \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} \). Its period is \( \pi \), and it is undefined at angles where \( \cos(\theta) = 0 \), such as \( \pi/2, 3\pi/2, \) etc.

In the unit circle, the tangent function is positive in the first and third quadrants and negative in the second and fourth quadrants. This is why \( \tan(2\pi/3) \) results in a positive value since \( 2\pi/3 \) lies in the second quadrant.

Co-function Identity

Co-function identities help find the values of trigonometric functions for complementary angles. For the tangent function, the co-function identity is \( \tan(\pi/2 - \theta) = \cot(\theta) \). This is useful when dealing with angles that are complements of familiar angles.

In our calculation, we used the identity \( \tan(2\pi/3) = \tan(\pi/2 - \pi/3) = \cot(\pi/3) \). This step reimagines \( 2\pi/3 \) in terms of \( \pi/2 - \theta \), making it easier to evaluate using known trigonometric values.

Exact Trigonometric Values

Exact trigonometric values are the known values of trigonometric functions at specific angles. For example, \( \tan(\pi/3) = \sqrt{3} \) and \( \cot(\pi/3) = 1/\sqrt{3} \). These values are important for finding expressions without a calculator.

When we determined \( \tan^{-1}(\tan(-4\pi/3)) \), we eventually evaluated \( \tan^{-1}(1/\sqrt{3}) \) to find the angle \( \pi/6 \) because \( \tan(\pi/6) \) is known to be \( 1/\sqrt{3} \). Knowing these exact values simplifies problems involving inverse trigonometric functions.